# Solving 1st and 2nd Order Linear ODEs with Constant Coefficients

In this post, I will cover the following:

*   First-Order Linear Differential Equations (ODEs)
    
*   Second-Order Linear Differential Equations (ODEs)
    

## First-Order Linear Differential Equations (ODEs)

* * *

First, we'll look at first-order linear differential equations with constant coefficients, covering the homogeneous case where the right-hand side is zero, followed by the non-homogeneous case with a trigonometric term on the right-hand side.

### Step-by-Step Solution (Homogeneous)

* * *

**Step 1. Separation of Variables**

Move the terms involving \\(y\\) to the left and terms involving \\(t\\) to the right (assuming \\(y \neq 0\\)):

$$\frac{dy}{dt}+Py = 0 \longrightarrow \frac{dy}{dt} = -Py \longrightarrow \frac{dy}{y} = -Pdt$$

*(Note: If* \\(y = 0\\)*, it trivially satisfies the equation, which is known as the trivial solution.)*

***Step 2. Integration***

Integrate both sides with respect to their variables, combining the constants of integration into a single constant \\(c\\) on the right:

$$\int\frac{dy}{y} = \int-Pdt \longrightarrow \ln|y| = -Pt + c$$

**Step 3. Exponentiation and Solving for** \\(y\\)

Exponentiate both sides with base $e$ to eliminate the natural logarithm:

$$e^{\ln|y|} = e^{-Pt+c} \longrightarrow |y| = e^ce^{-Pt}$$

Removing the absolute value gives:

$$y = \pm e^c e^{-Pt}$$

Since \\(\pm e^c\\) is simply an arbitrary non-zero constant, we can replace it with a single constant \\(C\\). Allowing \\(C = 0\\) accounts for the trivial solution \\(y = 0\\), giving the general solution:

$$y = C e^{-Pt}$$

### Step-by-Step Solution (Non-homogeneous)

* * *

**Step 1. Problem Setup and Euler's Formula**

$$\frac{dy}{dt} + Py = \cos(\omega t)$$

Before diving into the solution, recall **Euler's formula**, which connects complex exponentials with trigonometric functions:

$$e^{i\omega t} = \cos(\omega t) + i\sin(\omega t)$$

From this relation, cosine and sine can be represented as exponential forms:

$$\cos(\omega t) = \frac{e^{i\omega t} + e^{-i\omega t}}{2} \qquad \sin(\omega t) = \frac{e^{i\omega t} - e^{-i\omega t}}{2i}$$

**Step 2. Trigonometric Synthesis Setup**

![](https://cdn.hashnode.com/uploads/covers/697a014233686646a792dd2d/94554967-e71c-4944-ba6a-54e5fb530adb.png align="center")

From this geometric relationship, the trigonometric components and the phase shift \\(\alpha\\) are defined as:

$$\cos\alpha = \frac{P}{\sqrt{P^2 + \omega^2}} \qquad \sin\alpha = \frac{\omega}{\sqrt{P^2 + \omega^2}} \qquad \alpha = \arctan\left(\frac{\omega}{P}\right)$$

**Step 3: Variation of Parameters**

we replace the constant \\(C\\) with an unknown time-dependent function \\(C(t)\\)

$$y(t) = C(t) e^{-Pt}$$

Using the product rule, the first derivative of \\(y(t)\\) is:

$$\frac{dy}{dt} = \frac{dC}{dt} e^{-Pt} - P C(t) e^{-Pt}$$

Substitute into the differential equation:

$$\left( \frac{dC}{dt} e^{-Pt} - P C(t) e^{-Pt} \right) + P \Big( C(t) e^{-Pt} \Big) = \cos(\omega t)$$

The terms involving \\(PC(t)e^{-Pt}\\) cancel each other out:

$$\frac{dC}{dt} e^{-Pt} = \cos(\omega t) \longrightarrow \frac{dC}{dt} = e^{Pt}\cos(\omega t)$$

**Step 4. Solving for** \\(C(t)\\)

Integrating both sides with respect to \\(t\\) yields:

$$C(t) = \int e^{Pt}\cos(\omega t)\,dt = \frac{e^{Pt}}{\sqrt{P^2 + \omega^2}}\cos(\omega t - \alpha) + C_1$$

(Here, the harmonic addition formula was applied with phase angle α)

Finally, substituting \\(C(t)\\) back into \\(y(t) = C(t)e^{-Pt}\\):

$$y(t) = \left( \frac{e^{Pt}}{\sqrt{P^2 + \omega^2}}\cos(\omega t - \alpha) + C_1 \right) e^{-Pt} = \frac{1}{\sqrt{P^2 + \omega^2}}\cos(\omega t - \alpha) + C_1 e^{-Pt}$$

## Second-Order Linear Differential Equations (ODEs)

* * *

Next, we will cover second-order linear differential equations with constant coefficients, focusing on solving the homogeneous case.

$$\frac{d^2 y}{dt^2} + P_1 \frac{dy}{dt} + P_0 y = 0$$

### Step-by-Step Solution (homogeneous)

* * *

**Step 1. Form the Characteristic Equation**

Assume a trial solution of exponential form:

$$y = e^{at}$$

Differentiating yields \\(\frac{dy}{dt} = ae^{at}\\) and \\(\frac{d^2y}{dt^2} = a^2e^{at}\\). Substituting these into the differential equation gives:

$$a^2 e^{at} + P_1 a e^{at} + P_0 e^{at} = 0$$

Dividing through by \\(e^{at} \neq 0\\), we obtain the characteristic equation:

$$a^2 + P_1 a + P_0 = 0$$

**Step 2. Solve for the Roots**

Using the quadratic formula, solve for the roots \\(a_1\\), \\(a_2\\):

$$a = \frac{-P_1 \pm \sqrt{P_1^2 - 4P_0}}{2}$$

**Step 3. Determine the General Solution**

Please note that in this post, we will not cover the case where \\(P_1^2 - 4P_0 < 0\\)(complex conjugate roots).

*   Case 1: Two Distinct Real Roots (\\(P_1^2 - 4P_0 > 0\\))
    
    *   General Solution:
        

$$y = C_1 e^{a_1t} + C_2 e^{a_2t}$$

*   Case 2: Repeated Real Root (\\(P_1^2 - 4P_0 = 0\\))
    
    *   Since a repeated root yields only \\(e^{at}\\), we apply the reduction of order method to find the second linearly independent solution, which results in \\(te^{at}\\).
        
    *   General Solution:
        

$$y = C_1e^{at} + C_2 te^{at}$$

Now, by substituting the general solution \\(y = C_1Y + C_2Z\\) back into the original differential equation, we can see that it satisfies the equation as follows:

$$\frac{d^2}{dt^2}(C_1 Y + C_2 Z) + P_1 \frac{d}{dt}(C_1 Y + C_2 Z) + P_0 (C_1 Y + C_2 Z)$$

$$= C_1 \underbrace{\left(\frac{d^2 Y}{dt^2} + P_1 \frac{dY}{dt} + P_0 Y\right)}_{= 0} + C_2 \underbrace{\left(\frac{d^2 Z}{dt^2} + P_1 \frac{dZ}{dt} + P_0 Z\right)}_{= 0} = 0$$